来自匿名函数的PHP访问变量

Question

可能重复:
在创建匿名PHP函数期间呈现变量

我对PHP仍然很新,这让我感到困扰:

class Controller {
    ...
    ...
    function _activateCar() {
        $car_id = $this->data['car']->getId();
        // $car_id == 1
        $active_car = array_filter($this->data['cars'], function($car){
            // $car_id undefined
            return $car->getId() == $car_id;
        });
    }
    ...
    ...
}

为什么array_filter里面的函数不能访问$car_id变量？继续说未定义.

有没有其他方法可以$car_id访问而不是制作$_GET['car_id'] = $car_id;？使用global关键字没有帮助.

Answer 1

您需要添加use($car_id)到您的匿名函数,如下所示:

$active_car = array_filter($this->data['cars'], function($car) use($car_id){
    // $car_id undefined
    return $car->getId() == $car_id;
});

Answer 2

匿名函数可以使用use关键字导入选择变量:

$active_car = array_fiter($this->data['cars'],function($car) use ($car_id) {
    return $car->getId() == $car_id;
});